I am trying to prove the following statement
Let $A \subset \mathbb{R}^n$ be a bounded set that has volume. Let $f : A \mapsto \mathbb{R}, f \ge 0$ be an unbounded function. Suppose $C_i$ is a sequence of compact sets with volume, $C_i \subset A$ with $C_i$ increasing to $A$. Then $f$ is Riemann integrable iff $f$ is Riemann integrable on each $C_i$ and $\lim_{i \to \infty} \int_{C_i} f$ exists.
My attempt. Let $M > 0$ be given.
Define
$$
f_M(x) = \min \{ f(x), M\}
$$
$$
f_i = f \chi_{C_i}
$$
Then $\{f_i\}$ is a sequence of non-negative functions such that $f_i \le f_{i+1}$ and $f_i \to f$ pointwise. Now, if $f$ is integrable on each $C_i$ and $\lim_{i\to \infty} \int_{C_i} f < \infty$, then the set of discontinuities of $f_M$ on $C_i$ has measure zero for all $i$. Let $x_0 \in \{ \text{discontinuities of} \ f_M \ \text{on $A$} \}$ be given. Since $C_i$ increases to $A$, there exists $N$ such that $f_M$ is discontinuous at $x_0$ on $C_N$. So $x_0 \in \bigcup_{i=1}^\infty \{ \text{discontinuities of} \ f_M \ \text{on $C_i$} \ \}$. Since $\bigcup_{i=1}^\infty \{ \text{discontinuities of} \ f_M \ \text{on $C_i$} \ \}$ has measure zero, the set of discontinuities of $f_M$ on $A$ has measure zero. Thus, $f_M$ is integrable. Since this is true for all $M > 0$, $f$ is integrable. Then by Lebesgue's monotone convergence theorem,
$$
\lim_{i\to \infty} \int_{C_i} f = \lim_{i\to \infty} \int_A f_i = \int_A f
$$
Conversely, suppose $f$ is integrable. Since each $C_i$ is a compact subset of $A$ with volume, then $\{\text{discontinuities of} \ f_i \} \subset \partial C_i \cup \{ \text{discontinuities of} \ f \}$ has measure zero, so $f_i$ is integrable, i.e. $f$ is integrable on each $C_i$. Then by Lebesgue's monotone convergence theorem,
$$
\int_A f=\lim_{i\to\infty} \int_A f_i = \lim_{i\to\infty} \int_{C_i} f
$$.
Is my proof mathematically correct? Is there a more direct way of proving the statement? Any comment would be appriciated.
Best Answer
For the reverse implication, suppose $D \subset A$ is any compact set with volume (i.e., rectifiable). Since $D \subset \cup_{i=1}^\infty C_i=A$, there exists by compactness a set $C_m$ such that $D \subset C_m$.
Since $f$ is Riemann integrable on $C_m$ , it is Riemann integrable on $D$. (Here you would make the argument that the discontiuities of $f\chi_D$ must be a set of measure zero.) Thus, since $f \geqslant 0$,
$$\int_D f \leqslant \int_{C_m} f \leqslant \lim_{i\to \infty}\int_{C_i} f < \infty,$$
and
$$\int_A f := \sup_{D \subset \mathcal{D}} \int _D f\leqslant \lim_{i\to \infty}\int_{C_i} f,$$
where the supremum is taken over all compact rectifiable subsets of $A$. This proves the existence of the improper integral of $f$ over $A$ -- as it is usually defined.